Projectile Motion - object that is shot at an angle of 55 degrees

[Yellowkies_3275]

Homework Statement

Projectile motion
I have an an object that is shot at an angle of 55 degree's. (the velocity listed next to it was 35 m/s, which was not listed as V_o but I think I have to assume it is?) It is launched from the ground. it is meant to land to hit a button that is elevated by 32 feet. we have to find the distance away from the button to launch the object in order to hit the button.

Homework Equations

Delta x= Vo * t
delta y = 1/2gt^2 + Vo * t

The Attempt at a Solution

I tried a bunch of things that did not help me come to any sort of conclusion about the distance (x)

 

[gneill]

Please show us what you've tried.

 

[Charles Link]

For the $x=v_x t$, you need to have $v_x=v_o \cos(55^{\circ})$. Do you see why? $\\$ For the y equation, you need $y=-\frac{gt^2}{2}+v_{yi} t$ where $v_{yi}=v_o \sin(55^{\circ})$. The acceleration due to gravity is downward=thereby the minus sign. $\\$ Meanwhile, an important part you are missing is to separate the initial velocity $v_o$ at an angle of $55^{\circ}$ into horizontal and vertical components. I have shown you this result in the equations above. $\\$ Now that I got you started, can you figure out what to do to answer the question they are asking? Here's a hint: Can you solve for $y$ as a function of $x$? $\\$ Presently, the equations you have go through the point $x=0,y=0$ at $t=0$. What is $x$ when $y=32$ ? You need to determine the equation that gives you $y$ as a function of $x$, and then you can answer that question.

 

[Yellowkies_3275]

Please show us what you've tried.

sorry i didn't fill it out properly. i think i have a solution to the question now though, so i was wondeering if this thread could be deleted

 

[gneill]

sorry i didn't fill it out properly. i think i have a solution to the question now though, so i was wondeering if this thread could be deleted

After a thread has received replies, particularly where contributors have put in significant effort, our policy is to not remove the thread so that other members can benefit from the information.

 

[Ray Vickson]

Homework Statement

Projectile motion
I have an an object that is shot at an angle of 55 degree's. (the velocity listed next to it was 35 m/s, which was not listed as V_o but I think I have to assume it is?) It is launched from the ground. it is meant to land to hit a button that is elevated by 32 feet. we have to find the distance away from the button to launch the object in order to hit the button.

Homework Equations

Delta x= Vo * t
delta y = 1/2gt^2 + Vo * t

The Attempt at a Solution

I tried a bunch of things that did not help me come to any sort of conclusion about the distance (x)

Be very careful. If $y$ measures height above the ground, your formula $\Delta y = v_0 t + \frac{1}{2} g t^2$ will have the object accelerating rapidly upward, until it goes into outer space (because, conventionally, $g$ is a positive constant).

 

[Yellowkies_3275]

Be very careful. If $y$ measures height above the ground, your formula $\Delta y = v_0 t + \frac{1}{2} g t^2$ will have the object accelerating rapidly upward, until it goes into outer space (because, conventionally, $g$ is a positive constant).

How do I change the equation to keep it from accelerating into outer space?

 

[Charles Link]

How do I change the equation to keep it from accelerating into outer space?

You put a minus sign in front of the $\frac{g \, t^2}{2}$ term. $g=+9.8$ m/sec^2, but $a=-9.8$ m/sec^2. $\\$ Otherwise, your equation for $y$ has the form $y=At^2+Bt$ with $A>0$ , so that $y$ will get increasingly larger with time $t$.

 

[Yellowkies_3275]

You put a minus sign in front of the $\frac{g \, t^2}{2}$ term. $g=+9.8$ m/sec^2, but $a=-9.8$ m/sec^2. $\\$ Otherwise, your equation for $y$ has the form $y=At^2+Bt$ with $A>0$ , so that $y$ will get increasingly larger with time $t$.

You are a life, saver thank you very much for your explanations